AP EAMCET · PHYSICS · Work Power Energy
A ball is falling freely from a height. When it reaches \(10 \mathrm{~m}\) height from the ground its velocity is \(v_0\). It collides with the ground and loses \(50 \%\) of its energy and rises back to height of \(10 \mathrm{~m}\). Then the velocity \(v_0\) is
- A \(7 \mathrm{~m} / \mathrm{s}\)
- B \(10 \mathrm{~m} / \mathrm{s}\)
- C \(14 \mathrm{~m} / \mathrm{s}\)
- D \(16 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(14 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
At height \(10 \mathrm{~m}\), the kinetic energy is equal to potential energy then \(\frac{1}{2} m v_0^2=m g h\) \(\frac{1}{2} v_0^2=9.8 \times 10\) \(v_0^2=2 \times 9.8 \times 10\) \(v_0=\sqrt{196}\) \(v_0=14 \mathrm{~m} / \mathrm{s}\)
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